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Force between capacitor plates, partially filled with dielectric

  • Thread starter erst
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  • #1
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So for just vacuum in between one could say F=q*E = (σA)*(σ/2ε0) = A*σ^2 / (2ε0)

Alternatively, F = dU/dx, where x is the separation. U = 1/2 CV^2 = 1/2 QV = x*A*σ^2 / (2ε0). To get V, we just add up pieces of E-field*distance.

But now let's say part of x is a dielectric of thickness t and ε = εr*ε0, so vacuum is of thickness x-t.

Unfortunately, I get the exact same answer with the F = dU/dx approach since U = (A*σ^2 / (2ε0)) * (x - t + t/εr).

And with F=q*E, I'm not even sure what to do. If the dielectric was "suspended", the E-field would change right back to the vacuum value after passing through the dielectric so it's like it's not even there. Or if the dielectric is right at the plate's boundary, it would feel the reduced E-field.

I'm getting something wrong conceptually. I feel like there should be a difference in the force between the plates if there's a dielectric in there...
 

Answers and Replies

  • #2
ehild
Homework Helper
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You got an expression for the energy of the capacitor as function of x and t.
The force is -dU/dx, but it is different when t stays constant from that when t changes with x, that is the dielectrics moves in or out when you move a plane of the capacitor. The plates "feel" different electric forces, as it is the electric field that exerts force on the charges on the plate. The force on unit charge on a plate is proportional to the electric field, and it is different in the dielectrics.

ehild
 

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